1. MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = mol / L Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water. First calculate the moles of solute: 1.5 g NaCl ( 1 mole NaCl ) = 0.0257 moles of NaCl 58.5 g NaCl Next convert mL to L: 0.500 L of solution Last, plug the appropriate values into the correct variables in the equation: M = mol/L = 0.0257 moles / 0.500 L = 0.051 mol/L

2. How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution?

3. MOLARITY M = mol / L What is the molarity of hydroiodic acid if the solution is 47.0% HI by mass and has a density of 1.50 g/mL? First calculate the mass of solute in the 47.0% solution using the density. The 1.50 g/mL is the density of the solution but only 47.0% of the solution is the solute therefore: 47.0% of 1.50 g/mL = (0.470) (1.50 g/mL) = 0.705 g/mL density of solute Since molarity is given in moles per liter and not grams we must convert the g/mL to mol/mL using the molar mass. 0.705 g/mL (1 mole/ 128 g) = 0.00551 mol/mL Next convert mL to L: 0.00551 mol/mL (1000 mL/ 1L) = 5.51 mol/L = 5.51 M

4. MOLARITY & DILUTION M 1 V 1 = M 2 V 2 The act of diluting a solution is to simply add more water (the solvent) thus leaving the amount of solute unchanged. Since the amount or moles of solute before dilution and the moles of solute after the dilution are the same: mol b = mol a And the moles for any solution can be: moles=MV A relationship can be established such that M b V b = mol b = mol a = M a V a Or simply : M b V b = M a V a

5. MOLARITY & DILUTION Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M 1 = 0.05 mol/LM 2 = ? V 1 = 25.0 mLV 2 = 50.0 + 25.0 = 75.0 mL M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V 2 75.0 mL

6. MOLARITY & DILUTION Given a 6.00 M HCl solution, how would you prepare 250.0 mL of 0.150 M HCl? M 1 = 6.00 mol/LM 2 = 0.150 V 1 = ? mLV 2 = 250.0 mL M 1 V 1 = M 2 V 2 M 2 V 2 = V 1 = (0.150 mol/L) (250.0 mL) = 6.25 mL of 6 M HCl M 1 6.00 mol/L You would need 6.25 mL of the 6.00 M HCl reagent which would be added to about 100 mL of DI water in a 250.0 mL graduated cylinder then more water would be added to the mixture until the bottom of the menicus is at 250.0 mL. Mix well.

7. Ions in Solution Many chemicals are ionic. When dissolved in water they dissociate into their respective ion concentrations. [ion]=ion concentration Because BaCl 2 is ionic and dissociates: BaCl 2 (s)  Ba +2 (aq) + 2 Cl -1 (aq) Therefore the [ion] may differ as shown below:.35M BaCl 2 is really.35M = [Ba +2 ] and.70M = [Cl -1 ]

8. Using the Solubility Chart Some ions in solution are totally soluble, which means unlimited amounts can be present in solution together. Other ions have limited solubility together. If ions are found on the Solubility Chart: negative ion + positive ion  NOT SOL. means that the two are not soluble(will not dissolve) or will form a solid Precipitate (PPT) and fall to the bottom when mixed reducing or removing the ions from solution

9. Mixing Solutions Mix: 200ml of.5M Pb(NO 3 ) 2 solution with 500ml of.3M NaI solution. Find the [ion] of all remaining ions and the mass of the precipitate after mixing the two solutions.

10. THINK MOLES! Must find moles first

11. MAUL Chart (needed because moles are changing) Pb +2 + 2 Cl -1  PbCl 2(s) Continued next page MolesPb +2 (1)(2) Cl -1 Availiable.1.15 Used up.075.15 Left over.0250 Excess chemical Depleted

12. [Ion] after mixing & reacting